Prev: Meami.org Publishes Polynomial Time Quicksort Algortihm:
Next: Methods of Proving all of Incompleteness in Logic in Trivially Short Proofs & a Challenge
From: Transfer Principle on 30 May 2010 01:17 On May 29, 10:07 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > Earlier, Herc provided the following link to his proof: Oops, Google error. I wanted to write the link to Herc's website, where he attempted to prove that his claim holds, at least for finite lists. But apparently, Google fails whenever I try to provide the link or cut and paste the lists of reals from his website. (I'm not sure whether a full newsreader would have been able to handle my cutting and pasting.) Suffice it to say that Cooper's website failed to reorder the random list, leaving out one of the reals when it didn't match the diagonal. So Herc's claim, at least for the finite list, is false. We were not able to reorder the list from step #1 to obtain the diagonal chosen in step #2. One would think that it would be easy to find a reordering that matches the diagonal, considering that there are 60! reorderings yet only 3^60 diagonals. (60! is 53 orders of magnitude greater than 3^60.) And yet Cooper's algorithm failed to find a reordering for this diagonal. Note that this counterexample in the _finite_ case doesn't necessary disprove the _infinite_ claim that given a ternary real, there exists with probability 1 a reordering of the list of computable reals whose diagonal is the given real. So at least this conjecture is still open. Still, even if the conjecture is true, Herc is still indefensible because the conjecture doesn't imply that the reals are countable. > [You have a] terrible taste in people. There's that accusation again. I have a terrible taste in people merely because I've decided to defend posters like Cooper rather than posters like Spight. I defend posters who in my opinion need defending. I don't defend those like Spight, who often calls those who criticize ZFC wrong, since the main posters who attack them are the ones being called wrong. The posters who are called wrong are the ones who need defending, because they are being attacked left and right for being wrong. And even though Herc/Cooper turned out to be not worth defending in the end, that still doesn't mean that I'm going to start defending Spight. If the only way to have a good taste in people is to defend posters like Spight, then I'd much rather have a bad taste in people. Meanwhile, I noticed that a newbie poster has started a new thread attacking Cantor a few hours ago. I just posted in that thread right now. I hope to find out which of the four main cases describes that poster.
From: Transfer Principle on 30 May 2010 02:00 On May 29, 7:58 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Cantor's second proof of 'uncountable infinity' is based on trying to enumerate the powerset of naturals. > Here's my equivalent proof of uncountable infinity. > Let's assume an enumeration of naturals exists, call it N. Herc/Cooper is hardly the first poster on sci.math to attempt to apply the proof of the uncountability of the set of reals to the set of naturals. (In essence, one is attempting to prove that R is uncountable if and only if N is uncountable.) In previous threads, the proof attempt would entail listing the naturals and then diagonalizing. But this would typically produce a number with infinitely many digits on the diagonal. But standard naturals have only finitely many digits. Thus, the proof attempt is considered invalid in standard theory. But what about Herc's proof attempt? Let's see: > N= { > 1, > 2, > 3, > 4, > ... > } > Let's calculate a new natural MAX+1. > That is 4+1 = 5 > in this finite subset example. > Voila 5 is a new number not in N So Herc lists the elements of N. But then one wonders, what is the fifth element of Herc's list, given that its first four elements are 1,2,3, and 4? > Therefore no matter how big N is there is always a new element > that can be listed and therefore the size of the set N is bigger than infinity. If Cooper is assuming that N is the set {1,2,3,4}, then he has just proved that card(N) is greater than _four_, not countable infinity (aleph_0). But of course, standard theory does not dispute that card(N) > 4. We notice the similarity between Herc's proof and the Euclid's proof of the infinitude of the primes -- in each case, we take a finite list of elements of a set and use those elements to produce a new element, simply by adding one to either the product (in Euclid's case) or maximum (in Herc's case). But as with Euclid, this only shows that the set is _infinite_, not _uncountable_. Note that only _finite_ sets of naturals have maxima, so Herc is implying that the set is finite when he mentions MAX. But let's give Cooper the benefit of the doubt, since he does use an ellipsis, that he does intend his list to be infinite after all. Perhaps he intended to give the list: N = {1, 2, 3, 4, 6, 7, 8, 9, .... } so this list is infinite, yet we can find a natural number, namely 5, that's not on the list. Yet this still doesn't establish that N is uncountable. Before I proceed, one might wonder why I'm going through all this effort just to prove Herc wrong, since the other thread already established that Herc is in Case 3 (the case where we can call him wrong). Well, Herc did ask the question: > To eliminate expected confusion I will make my motives explicit, > the second proof is a spoof of Cantor's proof, if you can find the > flaw in my proof then attacked Spight for failing to answer it: > Your [Spight's] evasion of showing a distinction between my spoof > proof and Cantor's proof is noted So I'm trying to show the distinction between Herc's and Cantor's proof, as Cooper requests of us. To see where this lack of symmetry comes from, we notice the standard definition of countable: A set x is countable iff there exists a function f: omega -> x such that f is surjective. (Note: some mathematicians prefer "bijective" to "surjective" in this above definition, thereby excluding finite sets from the countable sets.) Thus, to find the definition of uncountable, we take the negation of the definition of countable: A set x is uncountable iff for every function f: omega -> x, f is not surjective. This uses the well-known fact from standard FOL that the negation of a formula with an existential quantifier is a formula with a universal quantifier. And now we see the lack of symmetry here. In order to show that a set is countable, we only need to find _one_ list that is complete, but to show that a set is uncountable, we must show that _every_ list is incomplete. So this is the difference between Cooper and Cantor. Cooper found only _one_ list of naturals that is missing an element, while Cantor showed that _every_ list of reals is missing at least one element. And furthermore, we can trivially find a list of naturals that isn't missing an element: N = {1, 2, 3, 4, 5, 6, 7, 8, 9, .... } From the definition of countable, this is all we need to establish countability. Recall that it only takes _one_ complete list to prove that a set is countable, yet we need _every_ list to be incomplete to prove that a set is uncountable. I hope that Herc won't take this post as bullying. Herc asked a question and I gave the answer, which is more than we can say of Spight, who evaded Herc's question.
From: Tim Little on 30 May 2010 03:48 On 2010-05-30, |-|ercules <radgray123(a)yahoo.com> wrote: > This is a serous attempt at a refutation of Cantor's powerset proof An attempt caused by mind-affecting drugs in your serum? Typo-jokes aside, it may have been a serious attempt, but also a serious failure. > The powerset proof is exactly this: > > Assume a large/infinite room full of boxes [...] No it isn't. At best that is an analogy for the proof, and not a very good one. > So extrapolate how that demonstrates higher infinities for me? The actual proof (not your abortive analogy) shows that the powerset of *any* set (infinite or not) is larger in cardinality than the original set. Since the set of natural numbers N is infinite, |2^N| is a larger infinity. - Tim
From: |-|ercules on 30 May 2010 05:53 "Transfer Principle" <lwalke3(a)lausd.net> wrote > On May 29, 7:58 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> Cantor's second proof of 'uncountable infinity' is based on trying to enumerate the powerset of naturals. >> Here's my equivalent proof of uncountable infinity. >> Let's assume an enumeration of naturals exists, call it N. > > Herc/Cooper is hardly the first poster on sci.math to > attempt to apply the proof of the uncountability of > the set of reals to the set of naturals. (In essence, > one is attempting to prove that R is uncountable if > and only if N is uncountable.) > > In previous threads, the proof attempt would entail > listing the naturals and then diagonalizing. But this > would typically produce a number with infinitely many > digits on the diagonal. But standard naturals have > only finitely many digits. Thus, the proof attempt is > considered invalid in standard theory. > > But what about Herc's proof attempt? Let's see: > >> N= { >> 1, >> 2, >> 3, >> 4, >> ... >> } >> Let's calculate a new natural MAX+1. >> That is 4+1 = 5 >> in this finite subset example. >> Voila 5 is a new number not in N > > So Herc lists the elements of N. But then one wonders, > what is the fifth element of Herc's list, given that its > first four elements are 1,2,3, and 4? > >> Therefore no matter how big N is there is always a new element >> that can be listed and therefore the size of the set N is bigger than infinity. > > If Cooper is assuming that N is the set {1,2,3,4}, then he > has just proved that card(N) is greater than _four_, not > countable infinity (aleph_0). But of course, standard > theory does not dispute that card(N) > 4. > > We notice the similarity between Herc's proof and the > Euclid's proof of the infinitude of the primes -- in each > case, we take a finite list of elements of a set and use > those elements to produce a new element, simply by adding > one to either the product (in Euclid's case) or maximum > (in Herc's case). But as with Euclid, this only shows > that the set is _infinite_, not _uncountable_. Note that > only _finite_ sets of naturals have maxima, so Herc is > implying that the set is finite when he mentions MAX. > > But let's give Cooper the benefit of the doubt, since he > does use an ellipsis, that he does intend his list to be > infinite after all. Perhaps he intended to give the list: > > N = > {1, > 2, > 3, > 4, > 6, > 7, > 8, > 9, > ... > } > > so this list is infinite, yet we can find a natural number, > namely 5, that's not on the list. Yet this still doesn't > establish that N is uncountable. > > Before I proceed, one might wonder why I'm going through all > this effort just to prove Herc wrong, since the other thread > already established that Herc is in Case 3 (the case where > we can call him wrong). Well, Herc did ask the question: > >> To eliminate expected confusion I will make my motives explicit, >> the second proof is a spoof of Cantor's proof, if you can find the >> flaw in my proof > > then attacked Spight for failing to answer it: > >> Your [Spight's] evasion of showing a distinction between my spoof >> proof and Cantor's proof is noted > > So I'm trying to show the distinction between Herc's and > Cantor's proof, as Cooper requests of us. > > To see where this lack of symmetry comes from, we notice the > standard definition of countable: > > A set x is countable iff there exists a function f: omega -> x > such that f is surjective. > > (Note: some mathematicians prefer "bijective" to "surjective" > in this above definition, thereby excluding finite sets from > the countable sets.) > > Thus, to find the definition of uncountable, we take the > negation of the definition of countable: > > A set x is uncountable iff for every function f: omega -> x, > f is not surjective. > > This uses the well-known fact from standard FOL that the > negation of a formula with an existential quantifier is a > formula with a universal quantifier. > > And now we see the lack of symmetry here. In order to show > that a set is countable, we only need to find _one_ list that > is complete, but to show that a set is uncountable, we must > show that _every_ list is incomplete. > > So this is the difference between Cooper and Cantor. Cooper > found only _one_ list of naturals that is missing an element, > while Cantor showed that _every_ list of reals is missing at > least one element. And furthermore, we can trivially find a > list of naturals that isn't missing an element: > > N = > {1, > 2, > 3, > 4, > 5, > 6, > 7, > 8, > 9, > ... > } > > From the definition of countable, this is all we need to > establish countability. Recall that it only takes _one_ > complete list to prove that a set is countable, yet we > need _every_ list to be incomplete to prove that a set > is uncountable. > > I hope that Herc won't take this post as bullying. Herc > asked a question and I gave the answer, which is more > than we can say of Spight, who evaded Herc's question. If you take every possible computer program and input every possible natural input then each program can be considered an element of the powerset of N. e.g. UTM(3,1) = 0 UTM(3,2) = 5 UTM(3,3) = 1 UTM(3,4) = 2 .... P(N)_3 = {0,5,1,2...} This computable powerset covers a LOT of subsets of N. You can't list a subset that it doesn't cover. Yet you all believe there are INFINITELY MORE missing subsets than present subsets! Back to the Cantor Spoof Proof... The fact you can find a complete set of N doesn't refute my demonstration, merely the proof itself. Restrict your deductions to using a subset of mathematics where you don't know N is missing an element. Then Cantor's logical deduction sequence used to imply higher infinities is exactly the same as the spoof proof. Herc
From: |-|ercules on 30 May 2010 05:55
"Tim Little" <tim(a)little-possums.net> wrote ... > On 2010-05-30, |-|ercules <radgray123(a)yahoo.com> wrote: >> This is a serous attempt at a refutation of Cantor's powerset proof > > An attempt caused by mind-affecting drugs in your serum? > > Typo-jokes aside, it may have been a serious attempt, but also a > serious failure. > > >> The powerset proof is exactly this: >> >> Assume a large/infinite room full of boxes [...] > > No it isn't. At best that is an analogy for the proof, and not a very > good one. Yes it is. Substitute box for set and fridge magnet for natural, it's a valid version of Cantor's powerset proof of uncountable infinity. The real question you should ask yourself is why you don't accept such a silly proof? Herc |